我如何根据另一个 table 中的值创建重复记录

How can i create duplicate records based the value in another table

我的数据库中有两个 table,Work_Order table 这是存储工单信息的来源 table 我还有 Work_Schedule table 其中包含工作时间表,告诉生产车间的人们要建造什么、何时建造以及建造多少。

Work_Order Table looks like

Work order  ItemCode    Size          Qty    Qty_per_HR 
41051        600111    14L-16.1        55          10

我想根据每小时的数量在上面的工作订单 table 中复制上面的工作订单行,并自动创建一个工作计划程序,如下所示。

其中 TARGET = Work_Order.Qty/Work_Order.Qty_per_HR

Work_Schedule Table

Id      Start Date/Time        End Date/Time       Work Order       Work Center     TARGET   ACTUAL
1001    2019-07-22 7:00AM      2019-07-22 8:00AM       41051             1         10       
1001    2019-07-22 8:00AM      2019-07-22 9:00AM       41051             1         10    
1001    2019-07-22 9:00AM      2019-07-22 10:00AM      41051             1         10    
1001    22019-07-22 10:15AM    2019-07-22 11:00AM      41051             1         10    
1001    22019-07-22 11:00AM    2019-07-22 12:00PM      41051             1         10  
1001    2019-07-22 1:30PM      2019-07-22 2:30PM       41051             1         5      

我的计划是使用 AfterInsert 在用户工作订单创建重复项时立即触发。

Schedule windows

这就是你想要的吗?

CREATE TABLE T(
  WorkOrder INT,
  ItemCode INT,
  Size VARCHAR(25),
  Qty INT,
  QtyPerH INT
);

INSERT INTO T VALUES
(41051,        600111,    '14L-16.1',        55,          10),
(41052,        600112,    '14L-16.2',        55,          5);

SELECT T.*
FROM T CROSS APPLY
(
  SELECT 1 N
  FROM master..spt_values
  WHERE [Type] = 'P'
        AND
        [Number] < (T.Qty / T.QtyPerH)
) TT;

Returns:

+-----------+----------+----------+-----+---------+
| WorkOrder | ItemCode |   Size   | Qty | QtyPerH |
+-----------+----------+----------+-----+---------+
|     41051 |   600111 | 14L-16.1 |  55 |      10 |
|     41051 |   600111 | 14L-16.1 |  55 |      10 |
|     41051 |   600111 | 14L-16.1 |  55 |      10 |
|     41051 |   600111 | 14L-16.1 |  55 |      10 |
|     41051 |   600111 | 14L-16.1 |  55 |      10 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
|     41052 |   600112 | 14L-16.2 |  55 |       5 |
+-----------+----------+----------+-----+---------+

Demo

这对于递归 CTE 来说似乎很自然:

with cte as (
      select convert(datetime, '2019-07-22 7:00AM') as dt, workorder, 1 as workcenter, qtyperh as target,
             itemcode, size, (qty - qtyperh) as qty, qtyperh
      from t
      union all
      select dateadd(hour, 1, dt), workorder, workcenter,
             (case when qty > qtyperh then qtyperh else qty end) as target,
             itemcode, size, (qty - qtyperh), qtyperh
      from cte
      where qty > 0
     )
select cte.*,
       dateadd(second, 60 * 60 * target / qtyperh, dt) as end_dt
from cte
order by workorder, dt;

Here 是一个 db<>fiddle.